Solveeit Logo
Login

Question

Question: If \(\tan \left( A+B \right)=p\) and \(\tan \left( A-B \right)=q\) , show that \(\tan 2A=\dfrac{p+q}...

If tan⁡(A+B)=p\tan \left( A+B \right)=p and tan⁡(A−B)=q\tan \left( A-B \right)=q , show that tan⁡2A=p+q1−pq\tan 2A=\dfrac{p+q}{1-pq} .

Explanation

Solution

Hint: Substitute the value or p and q, and try to simplify the right-hand side of the equation that we need to prove by using the formula tan⁡(X+Y)=tanX⁡+tanY⁡1−tanYtanX⁡\tan \left( X+Y \right)=\dfrac{\operatorname{tanX}+\operatorname{tanY}}{1-tanY\operatorname{tanX}} .

Complete step-by-step answer:
Now we will start with the simplification of the right-hand side of the equation that is given in the question which we are asked to prove.
p+q1−pq\dfrac{p+q}{1-pq}
It is given in the question that p=tan(A+B) and q=tan(A-B). So, we will substitute the values of p and q in our expression, we get
tan⁡(A+B)+tan⁡(A−B)1−tan⁡(A+B)tan⁡(A−B)\dfrac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B)\tan (A-B)}
Now we know the formula that tan⁡(X+Y)=tanX⁡+tanY⁡1−tanYtanX⁡\tan \left( X+Y \right)=\dfrac{\operatorname{tanX}+\operatorname{tanY}}{1-tanY\operatorname{tanX}} . On using this in our expression, the expression becomes
=tan⁡(A+B+(A−B))=\tan \left( A+B+\left( A-B \right) \right)
=tan⁡(A+B+A−B)=\tan \left( A+B+A-B \right)
=tan⁡2A=\tan 2A
The left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we can say that we have proved the equation tan⁡2A=p+q1−pq\tan 2A=\dfrac{p+q}{1-pq} is proved.

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, be careful about the signs in the formula of tan(A+B). Whenever you are dealing with an expression having cotangents, secant, and cosecant involved, it is better to convert it to an equivalent expression in terms of sine, cosine, and tangent, as most of the formulas we know are valid for sine, cosine, and tangents only.